Integrand size = 30, antiderivative size = 102 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\frac {c^2 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {2 c^2 \log (1+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
c^2*ln(cos(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1 /2)+2*c^2*ln(1+sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f* x+e))^(1/2)
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.60 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\frac {c^2 (\log (\cos (e+f x))+2 \log (1+\sec (e+f x))) \tan (e+f x)}{f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
(c^2*(Log[Cos[e + f*x]] + 2*Log[1 + Sec[e + f*x]])*Tan[e + f*x])/(f*Sqrt[a *(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4400, 27, 86, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a \sec (e+f x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4400 |
\(\displaystyle -\frac {a c \tan (e+f x) \int \frac {c \cos (e+f x) (1-\sec (e+f x))}{a (\sec (e+f x)+1)}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {c^2 \tan (e+f x) \int \frac {\cos (e+f x) (1-\sec (e+f x))}{\sec (e+f x)+1}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {c^2 \tan (e+f x) \int \left (\cos (e+f x)-\frac {2}{\sec (e+f x)+1}\right )d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^2 \tan (e+f x) (\log (\sec (e+f x))-2 \log (\sec (e+f x)+1))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((c^2*(Log[Sec[e + f*x]] - 2*Log[1 + Sec[e + f*x]])*Tan[e + f*x])/(f*Sqrt [a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))
3.2.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a*c*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^(n - 1/2)/x), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Time = 2.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {\sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, c \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \ln \left (-\frac {4 \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right ) \left (\cot \left (f x +e \right )-\cos \left (f x +e \right ) \cot \left (f x +e \right )\right )}{f a \left (\cos \left (f x +e \right )-1\right )}\) | \(85\) |
risch | \(\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, x}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (f x +e \right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {4 i c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}+\frac {i c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\) | \(390\) |
-1/f/a*(-c*(sec(f*x+e)-1))^(1/2)*c*(a*(sec(f*x+e)+1))^(1/2)*ln(-4*cos(f*x+ e)/(cos(f*x+e)+1)^2)/(cos(f*x+e)-1)*(cot(f*x+e)-cos(f*x+e)*cot(f*x+e))
\[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\int { \frac {{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{\sqrt {a \sec \left (f x + e\right ) + a}} \,d x } \]
\[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}\, dx \]
Time = 0.37 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59 \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=-\frac {{\left ({\left (f x + e\right )} c + c \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 4 \, c \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right )\right )} \sqrt {c}}{\sqrt {a} f} \]
-((f*x + e)*c + c*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*c*ar ctan2(sin(f*x + e), cos(f*x + e) + 1))*sqrt(c)/(sqrt(a)*f)
Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Timed out. \[ \int \frac {(c-c \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]